Question

An orange juice producer buys oranges from a large orange grove that has one variety of orange. The amount of juice squeezed from these oranges is approximately normally distributed, with a mean of 5.0 ounces and a standard deviation of 0.40 ounce. Suppose that you select a sample of 25 oranges. The probability is 70% that the sample mean amount of juice will be contained between what two values symmetrically distributed around the population mean?

Answer 1

Answer:

The probability is 70% that the sample mean amount of juice will be contained between 4.9168 ounces and 5.0832 ounces.

Step-by-step explanation:

To solve this question, the Normal probability distribution and the Central Limit Theorem are important.

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$, a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 5, \sigma = 0.4, n = 25, s = \frac{0.4}{\sqrt{25}} = 0.08$

The probability is 70% that the sample mean amount of juice will be contained between what two values symmetrically distributed around the population mean?

The lower end of this interval is the value of X when Z has a pvalue of 0.5 - 0.7/2 = 0.15

The upper end of this interval is the value of X when Z has a pvalue of 0.5 + 0.7/2 = 0.85

Lower end

X when Z has a pvalue of 0.15. So X when $Z = -1.04$.

$Z = \frac{X - \mu}{\sigma}$

Due to the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$-1.04 = \frac{X - 5}{0.08}$

$X - 5 = -1.04*0.08$

$X = 4.9168$

Upper end

X when Z has a pvalue of 0.15. So X when $Z = 1.04$.

$Z = \frac{X - \mu}{\sigma}$

Due to the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$1.04 = \frac{X - 5}{0.08}$

$X - 5 = 1.04*0.08$

$X = 5.0832$

The probability is 70% that the sample mean amount of juice will be contained between 4.9168 ounces and 5.0832 ounces.